Question

When an object is placed 70.0 cm from a certain converging lens, it forms a real image. When the object is moved to 32.0 cm from the lens, the image moves 9.00 cm farther from the lens. Find the focal length of the lens.

Answer 1

as given:

first case: object distance u = -70.0cm ,

1/v- 1/u = 1/f or, 1/v + 1/70 = 1/f ..........................1)

second case: now object distnace = -(70+32) = -102 cm , v' = (v+9) cm

1/v' - 1/u = 1/f

or, 1/(v+9) + 1/102 = 1/f ........................................2)

equatting euqation 1) and 2) , we get

1/v + 1/70 = 1/(v+9) + 1/102

or, 1/v - 1/(v+9) = 1/102 - 1/70

or, (v+9-v)/(v^2 + 9v) = - 1/223.125

or, v^2 + 9v - 2008 = 0 (approx)

or v = 81 cm

so 1/f = 1/v - 1/ u = 1/81 - 1/-70 = 0.02663

so f = 37.54 cm ...................the focal length of the lens...............................and